大数乘法的分治策略C++实现

来源：互联网发布：教化妆的软件编辑：程序博客网时间：2024/06/08 23:10
最近学习Coursera上的算法课，刚开始接触分治算法（Divide & Conquer），第一周的测试题为两个大位整数的乘法，结合自己自学的弱爆的C++，各种查资料还调了一整天的Bug才写出来个不完整的东西。只怪自己平时代码敲得少，不熟练。很多bug遇到过就会对此很敏感，所以多去造轮子，多尝试自己去实现才是最快的学习方法。哈哈哈，一堆废话，自勉之~

下面是自己写的源码，其中还有许多不妥的地方，第一次写（捂脸跑~）完好兴奋呀。。。

#include <stdio.h>#include <iostream>#include <list>#include <iterator>#include <string>using namespace std;void PrintList(list<char>& m); //计算结果输出函数void PrintMenu();   //功能菜单list<char> TransForm(string &s); //转换函数list<char> BigAdd(list<char>&a, list<char>&b); //大数加法list<char> BigSub(list<char>&a, list<char>&b); //大数减法list<char> BigMul(list<char>&a, list<char>&b); //大数乘法void BigDev();   大数除法，没实现//list<char> result;  //设为全局变量,则可以返回其引用以节省内存开销，    但在递归时不知道怎么处理。。。int main(int argc, char *argv[]){    list<char> num1, num2;    list<char> result;    char choice;    string input1, input2;    while (true)    {        cout << "**********************************" << endl;        cout << "   大整数计算器 VER1.0    " << endl;        cout << "**********************************" << endl;        PrintMenu();        result.clear();        cin >> choice;        fflush(stdin);  //清空缓冲区，删去多余输入        switch (choice)        {        case '1':            cout << "请输入第一个数  " << endl;            cin >> input1;            num1 = TransForm(input1);            cout << "请输入第二个数  " << endl;            cin >> input2;            num2 = TransForm(input2);            result = BigAdd(num1, num2);            PrintList(num1);            cout << " + ";            PrintList(num2);            cout << " = ";            PrintList(result);//            cout << endl << endl;            break;        case '2':            cout << "请输入第一个数  " << endl;            cin >> input1;            num1 = TransForm(input1);            cout << "请输入第二个数  " << endl;            cin >> input2;            num2 = TransForm(input2);            result = BigSub(num1, num2);            PrintList(num1);            cout << " - ";            PrintList(num2);            cout << " = ";            PrintList(result);//            cout << endl << endl;            break;        case '3':            cout << "请输入第一个数  " << endl;            cin >> input1;            num1 = TransForm(input1);            cout << "请输入第二个数  " << endl;            cin >> input2;            num2 = TransForm(input2);            result = BigMul(num1, num2);            PrintList(num1);            cout << " * ";            PrintList(num2);            cout << " = ";            PrintList(result);//            cout << endl << endl;            break;        case '4':            return 0;        }    }    system("pause");    return 0;}list<char> BigAdd(list<char>& a, list<char>& b)  //大数加法{    int len1, len2;    int carry = 0;//carry为进位    list<char> result;    len1 = a.size(), len2 = b.size();    //补全位置    if (len1 <= len2)    {        for (int i = 0; i<len2 - len1; i++)        {            a.push_front('0');        }    }    else    {        for (int i = 0; i<len1 - len2; i++)        {            b.push_front('0');        }    }    auto iter1 = a.rbegin(), iter2 = b.rbegin();//从低位遍历，使用反序迭代器很方便    int i = 0;    for (; iter1 != a.rend(); ++iter1, ++iter2)    {        i = ((*iter1 - '0') + (*iter2 - '0')) + carry;        result.push_front((i % 10 + '0'));        carry = i / 10;    }    if (carry)    {        result.push_front(carry + '0');    }    return result;}list<char> BigSub(list<char>&a, list<char>&b)    //大数减法{    list<char> result;    int len1 = a.size(), len2 = b.size();    bool flag;       //判断a,b的大小    if (len1<len2)        flag = false;    else if (len1 == len2)    {        auto iter_1 = a.begin(), iter_2 = b.begin();        for (; iter_1 != a.end(); iter_1++, iter_2++)        {            if ((*iter_1)<(*iter_2))            {                flag = false;                break;            }            flag = true;        }    }    else        flag = true;    if (flag)  //a>b时    {        for (int i = 0; i<len1 - len2; i++)        {            b.push_front('0');  //补全位置        }        auto iter1 = a.rbegin(), iter2 = b.rbegin(); //大赞反向迭代器！        for (int j = 0; iter1 != a.rend(); )        //容器的遍历尽量使用其迭代器控制，不要用int i和a.size()来控制        {            if (((*iter1) - (*iter2))<0)            {                j = 10 + *iter1 - *iter2;                result.push_front(j + '0');                (*(++iter1))--, ++iter2;            }            else            {                j = *iter1 - *iter2;                result.push_front(j + '0');                ++iter1, ++iter2;            }        }        return result;    }    else     //a<b,先计算b-a,在结果前加'-'    {        for (int i = 0; i<len2 - len1; i++)        {            a.push_front('0');        }        auto iter1 = a.rbegin(), iter2 = b.rbegin();        for (int j = 0; iter1 != a.rend(); )        {            if ((*iter2 - *iter1)<0)            {                j = 10 + *iter2 - *iter1;                result.push_front(j + '0');                (*(++iter2))--, ++iter1;            }            else            {                j = *iter2 - *iter1;                result.push_front(j + '0');                ++iter2, ++iter1;            }        }        result.push_front('-');        return result;    }}list<char> BigMul(list<char> &a, list<char> &b)  //大数乘法{    list<char> AC, BD, A, B, C, D; //ABCD分别为两数字的高位段、低位段    list<char> result;    int k;    if ((a.size() == 1) && (b.size() == 1))    {        k = (*a.begin() - '0')*(*b.begin() - '0');  //k最小为0，最大为81        int i, j;        i = k % 10, j = k / 10;        result.push_front(i + '0');        if (j)          //如果k<10,还是不要进0了            result.push_front(j + '0');    }    else if ((a.size() == 1) && (b.size()>1))    {        int m = b.size() / 2;// n =(b.size()+1)/2;        auto iter = b.rbegin();        list<char> HighRes, LowRes;        for (int i = b.size(); i>m; i--)        {            D.push_front(*iter);            iter++;        }        for (; iter != b.rend();)        {            C.push_front(*iter);            iter++;        }        HighRes = BigMul(a, C);        for (int i = 0; i<(b.size() - m); i++)        {            HighRes.push_back('0');        }        LowRes = BigMul(a, D);        result = BigAdd(HighRes, LowRes);    }    else if ((b.size() == 1) && (a.size()>1))    {        int m = a.size() / 2;// n =(b.size()+1)/2;        auto iter = a.rbegin();        list<char> HighRes, LowRes;        for (int i = a.size(); i>m; i--)        {            B.push_front(*iter);            iter++;        }        for (; iter != a.rend();)        {            A.push_front(*iter);            iter++;        }        HighRes = BigMul(b, A);        for (int i = 0; i<(a.size() - m); i++)        {            HighRes.push_back('0');        }        LowRes = BigMul(b, B);        result = BigAdd(HighRes, LowRes);    }    else    {        list<char> A, C, B, D;        int m = a.size() / 2, n = b.size() / 2;        auto iter1 = a.rbegin(), iter2 = b.rbegin();        for (int i = a.size(); i>m; i--)        {            B.push_front(*iter1);  //写入num1低位分段            iter1++;        }        for (; iter1 != a.rend();)        {            A.push_front(*iter1);  //写入num1高位分段            iter1++;        }        for (int j = b.size(); j>n; j--)        {            D.push_front(*iter2);  //写入num2低位分段            iter2++;        }        for (; iter2 != b.rend();)        {            C.push_front(*iter2);  //写入num2高位分段            iter2++;        }        AC = BigMul(A, C);        for (int i = 0; i<(a.size() - m); i++)        {            AC.push_back('0');        }        for (int j = 0; j<(b.size() - n); j++)        {            AC.push_back('0');        }        BD = BigMul(B, D);        list<char> AD, BC;        AD = BigMul(A, D);   //其实还有一种算法可以少计算一次乘法:                                ```math                                (a*10^n+b)*(c*10^n+d) = ac*10^2n+bd+(ac+bd)*10^n                                其中 ac 和 bd共需要两次乘法，考虑 ac + bd = （a+b）*(c+d)-ac-bd, ac及bd上一步已求出，故可少做一次乘法!!!                                ```        ）        for (int i = 0; i<(a.size() - m); i++)        {            AD.push_back('0');  //补齐低位        }        BC = BigMul(B, C);        for (int j = 0; j<(b.size() - n); j++)        {            BC.push_back('0');        }        result = BigAdd(AC, BD);        result = BigAdd(result, AD);        result = BigAdd(result, BC);        //midres=BigSub( BigSub( BigMul( BigAdd(Num1High,Num1Low), BigAdd(Num2High,Num2Low)),BigMul(Num1High,Num2High) ), BigMul(Num1Low,Num2Low));        //result=BigAdd( BigAdd(BigMul(Num1High,Num2High),lowres), midres);        //这里由于函数返回类型并非 list <char>& 类型，所以不能将函数本身作为参数！        //在含有递归的函数里，如何才能返回引用类型呢？？？求解。。。    }    return result;}void PrintMenu(){    cout << endl;    cout << "请输入要进行的运算： " << endl;    cout << endl;    cout << "1.  " << "加法运算" << endl;    cout << "2.  " << "减法运算" << endl;    cout << "3.  " << "乘法运算" << endl;    cout << "4.  " << "退出" << endl;    cout << endl;}list<char> TransForm(string &s){    list<char> result;    for (int i = 0; i<s.length(); i++)    {        result.push_back(s[i]);    }    return result;}void PrintList(list<char>& m){    bool flag = false;  //记录数字前面是否有多余的0    list<char>::iterator iter = m.begin();    if (*iter == '-')    {        cout << '-';        m.pop_front();        iter = m.begin();    }    while (iter != m.end())    {        if ((*iter == '0') && !flag)        {            m.pop_front(); iter = m.begin(); //过滤掉多余的0            continue;        }        else        {            flag = true;            cout << *iter;            iter++;        }    }}
初始界面^_^
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